I am trying to prove the following functional equation:
$$ \frac{ζ(s)}{ζ(1 − s)} = −s \frac{\tan(\frac{πs}{2})ζ(1 + s)}{2πζ(−s)}, $$
where $\zeta(s)$ is the Riemann zeta function for $s\in\mathbb{C}$, and $\Gamma$ is the Gamma function.
I tried using Riemann's functional equation:
$$ ζ(s) = 2^s π^{s−1}\sin(\frac{πs}{2})Γ(1 − s)ζ(1 − s), $$
and by the substitution $s\to1-s$ quickly derived:
$$ ζ(1 − s) = 2 (2π)^{-s}\cos(\frac{πs}{2})Γ(s)ζ(s). $$
However, I am stuck at this point, as no matter how I combine these two equations I never get the functional equation in my question.
Any help would be appreciated.
We also know that $\zeta(-s) = -2^{-s}\pi^{-s-1}\sin (\frac{\pi s}{2}) \Gamma(1+s)\zeta(1+s)$ and $\zeta(1+s) = 2(2\pi)^s \cos(\frac{\pi s}{2})\Gamma(-s)\zeta(-s)$.
So, we have: $$\frac{\zeta(s)}{\zeta(1-s)} = \frac{1}{\pi} (2\pi)^s \sin(\frac{\pi s}{2})\Gamma(1-s) \tag{1}$$
We also know that by property of the gamma function: $\Gamma(c)c = \Gamma(1+c)$.
Thus, $\Gamma(1-s) = (-s)\Gamma(-s)$. So, $(1)$ becomes: $$\frac{1}{\pi}(-s)(2\pi)^s\sin (\frac{\pi s}{2}) \Gamma(-s) =-2s(2\pi)^{s-1}\sin(\frac{\pi s}{2})\Gamma(-s)$$ $$=-\frac{s}{2\pi} \tan(\frac{\pi s}{2}) \frac{\zeta(1+s)}{\zeta(-s)}$$